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If $A=\left\{x \in R / \frac{\pi}{4} \leq x \leq \frac{\pi}{3}\right\}$ and $f(x)=\sin x-x$, then $f(A)$ is equal to
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Verified Answer
The correct answer is:
$\left[\frac{\sqrt{3}}{2}-\frac{\pi}{3}, \frac{1}{\sqrt{2}}-\frac{\pi}{4}\right]$
Given, $f(x)=\sin x-x$
Here, $\quad \frac{\pi}{4} \leq x \leq \frac{\pi}{3}$
$$
f\left(\frac{\pi}{4}\right) \geq f(x) \geq f\left(\frac{\pi}{3}\right)
$$
$\left(\because f(x)\right.$ is decreasing function in $\left.x \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right]\right)$
$$
\begin{aligned}
& \sin \frac{\pi}{4}-\frac{\pi}{4} \geq f(x) \geq \sin \frac{\pi}{3}-\frac{\pi}{3} \\
& \frac{1}{\sqrt{2}}-\frac{\pi}{4} \geq f(x) \geq \frac{\sqrt{3}}{2}-\frac{\pi}{3} \\
& \therefore \quad f(A) \in\left[\frac{\sqrt{3}}{2}-\frac{\pi}{3} \cdot \frac{1}{\sqrt{2}}-\frac{\pi}{4}\right]
\end{aligned}
$$
Here, $\quad \frac{\pi}{4} \leq x \leq \frac{\pi}{3}$
$$
f\left(\frac{\pi}{4}\right) \geq f(x) \geq f\left(\frac{\pi}{3}\right)
$$
$\left(\because f(x)\right.$ is decreasing function in $\left.x \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right]\right)$
$$
\begin{aligned}
& \sin \frac{\pi}{4}-\frac{\pi}{4} \geq f(x) \geq \sin \frac{\pi}{3}-\frac{\pi}{3} \\
& \frac{1}{\sqrt{2}}-\frac{\pi}{4} \geq f(x) \geq \frac{\sqrt{3}}{2}-\frac{\pi}{3} \\
& \therefore \quad f(A) \in\left[\frac{\sqrt{3}}{2}-\frac{\pi}{3} \cdot \frac{1}{\sqrt{2}}-\frac{\pi}{4}\right]
\end{aligned}
$$
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